SEMI FINAL STAGE 2025
Mfantsipim School: 42 points
GSTS: 23 points
Mankranso SHS: 15 points
QUESTION
A reaction proceeds in two equilibrium steps.
Step 1: $A+B\rightleftharpoons C$, with equilibrium constant $K_1=140$.
Step 2: $2C\rightleftharpoons D$, with equilibrium constant $K_2=0.25$.
What is the equilibrium constant for the reaction $2A+2B\rightleftharpoons D$?
ANSWER: 4900.
SOLUTION:
Doubling Step 1 ($A+B\rightleftharpoons C$) gives $2A+2B\rightleftharpoons 2C$ with equilibrium constant
$K_1^2 = 140^2$
$K_1^2 = 19600$
(squaring a reaction squares its constant).
Adding this to Step 2 ($2C\rightleftharpoons D$, $K_2=0.25$) gives the overall reaction $2A+2B\rightleftharpoons D$.
When equilibrium steps are added, their constants multiply:
$K = K_1^2\times K_2$
$K = 19600\times0.25$
$K = 19600 \times 25\times10^{-2}$
$K = 4900$
---
PRACTICE QUESTIONS
1. QUESTION: First step: $A+B\rightleftharpoons C$, $K_1 = 12$. Second step: $2C\rightleftharpoons D$, $K_2 = 0.50$. What is the equilibrium constant for $2A+2B\rightleftharpoons D$?
ANSWER: 72 or $7.2\times10^{1}$
SOLUTION:
Doubling the first step squares its constant
$K = K_1^2\times K_2$
$K = 12^2\times0.50$
$K = 144\times0.50$
$K = 144 \times 5\times10^{-1}$
$K = 72$
2. QUESTION: First step: $X\rightleftharpoons 2Y$, $K_1 = 0.20$. Second step: $Y\rightleftharpoons Z$, $K_2 = 5.0$. What is the equilibrium constant for $X\rightleftharpoons 2Z$?
ANSWER: 5.0
SOLUTION:
The second step must be doubled, which squares its constant
$K = K_1\times K_2^2$
$K = 0.20\times5.0^2$
$K = 0.20\times25$
$K = 2\times10^{-1} \times 25$
$K = 5.0$
3. QUESTION: The equilibrium constant for $P\rightleftharpoons Q$ is 25. What is the equilibrium constant for $2Q\rightleftharpoons 2P$?
ANSWER: 0.0016 or $1.6\times10^{-3}$
SOLUTION:
Reversing a reaction inverts K; doubling it squares K
$K' = \left(\dfrac{1}{25}\right)^2$
$K' = \dfrac{1}{625}$
$K' = 1.6\times10^{-3}$
4. QUESTION: First step: $A\rightleftharpoons B$, $K_1 = 4.0$. Second step: $B\rightleftharpoons C$, $K_2 = 0.25$. What is the equilibrium constant for $A\rightleftharpoons C$?
ANSWER: 1.0
SOLUTION:
Adding two steps multiplies their constants
$K = K_1\times K_2$
$K = 4.0\times0.25$
$K = 4 \times 25\times10^{-2}$
$K = 1.0$
5. QUESTION: The equilibrium constant for $P\rightleftharpoons Q$ is 0.20. What is the equilibrium constant for $Q\rightleftharpoons P$?
ANSWER: 5.0
SOLUTION:
Reversing a reaction inverts K
$K' = \dfrac{1}{K}$
$K' = \dfrac{1}{0.20}$
$K' = 5.0$
6. QUESTION: The equilibrium constant for $X\rightleftharpoons Y$ is 3.0. What is the equilibrium constant for $2X\rightleftharpoons 2Y$?
ANSWER: 9.0
SOLUTION:
Doubling a reaction squares K
$K' = K^2$
$K' = 3.0^2$
$K' = 9.0$